- Dear friends,

First of all I wish you a very happy and fruitful 2005.

To start the new year let me offer a little theorem:

If A'B'C' is a circumcevian triangle, and A"B"C" are the reflections of

A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C), (A"BC")

and (AB"C") are concurrent in one point.

Kind regards,

Floor. - Dear Jean-Pierre,

[FvL]> > To start the new year let me offer a little theorem:

[JPE]

> > If A'B'C' is a circumcevian triangle, and A"B"C" are the

> reflections of

> > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),

> (A"BC")

> > and (AB"C") are concurrent in one point.

> Note that if A',B',C' lie on the circumcircle, the circles A"B"C,

If we have A' and B', we can construct A", B", and then (A"B"C), and after

> B"C"A, C"A"B have a common point.

> I suspect that this common point lies on the circumcircle if and

> only if the lines AA', BB', CC' concur.

> Any idea?

that the common point P of (A"B"C) and the circumcircle. That means we have

the circles (AB"P)=(AB"C") and (A"BP)=(A"BC") as well. The point C" is thus

fixed as the second intersection of these circles (apart from P). So only

one point C", and consequently one point C' satisfies the condition. This

must be the circum-cevian trace of the intersection of AA' and BB'.

Kind regards,

Floor.